Evaluate the definite integral. $\int^2_{0}\big(4x^3-3x^2+2x\big)\,dx = $
Explanation: First, use the power rule: $\int^2_{0}\big(4x^3-3x^2+2x\big)\,dx~=~x^4-x^3+x^2\Bigg|^2_{0}$ Second, plug in the limits of integration: $[2^4-2^3+2^2]-[0] = 12$. The answer: $\int^2_{0}\big(4x^3-3x^2+2x\big)\,dx = 12$